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The number of four-digit integers which are greater than 1000 and divisible by both 2 and 3, but not by 5, is

Solution

✅ Correct Option: 3

Number of items in an AP =l−ad+1= \dfrac{l-a}{d}+1, where:

l=l = last term

a=a = first term

d=d = common difference


Integers divisible by 22 and 3⇒3 \Rightarrow divisible by 66

Integers divisible by 6⇒1002,1008,…,99966 \Rightarrow 1002,1008, \ldots ,9996

n1=9996−10026+1n_1= \frac{9996-1002}{6}+1

n1=1500n_1=1500


Integers divisible by 2,32,3 and 5⇒5 \Rightarrow divisible by 3030

Integers divisible by 30⇒1020,1050,…,999030 \Rightarrow 1020,1050, \ldots, 9990

n2=9990−102030+1n_2 = \frac{9990-1020}{30}+1

n2=300n_2=300


∴\therefore Total integers divisible by 22 and 33 but not 55:

=n1−n2=1500−300=1200=n_1-n_2=1500-300 = 1200

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